[AnneBee]

Given the following reaction at a constant temperature:

CH₃COCH₃ + I₂ --> CH₃COCH₂I + HI

1  0.100 mol/L          0.100 mol/L          1.16 x 10^-7 mol/ls
2  0.0500mol/L         0.100 mol/L          5.79 x 10^-8
3  0.0500 mol/L        0.500 mol/L          5.78 x 10^-8

Determine the rate law for this reaction

HERE IS MY WORK:

Trial 1-->2 [CH3COCH3]^x = rate
                 0.0500/0.100 = 5.79 x 10^-8/1.16 x 10^-7 mol/ls
                 0.5x = 2.9
                 This is incorrect why?

Trial 2 -->3   [I2]x = rate
                    [0.500/100]^y=5.78 x 10^-8/ 5.79 x 10^-8
                     5y = 1
                     y = 0 
                     This is also wrong why?

Any help would be greatly appreciated!!

[MNIO]

where did you get
   [CH3COCH3]^x = rate
   0.0500/0.100 = 5.79 x 10-8/1.16 x 10-7 mol/ls     what happened to the ^x
   0.5x = 2.9                 ^x?, 2.9?
should be
   ([CH3COCH3#2] / [CH3COCH3#1])^x = (rate#2 / rate#1)
   (0.0500/0.100)^x = (5.79 x 10-8 / 1.16 x 10-7)
   0.5^x = 0.5
   x = 1

same thing for y, you wrote
  [I2]x = rate
  [0.500/100]y=5.78 x 10-8/ 5.79 x 10-8
   5y = 1
  y = 0
should be
  ([I2#3] / [I2#2])^x = rate#3 / rate#2
  ([0.500/0.100])^y = (5.78 x 10-8 / 5.79 x 10-8)
   5^y = 1
   y = 0

*****************
finishing the problem
*****************
now we know
   rate = k * [CH3CO2H]¹ * [I2]⁰
or simply
   rate = k *  [CH3CO2H]

now we need to plug in any data point and solve for "k"
  k = rate / [CH3CO2H] = (1.16e-7 M/s) / (0.100 M) = 1.16e-6 / sec

and now we can write the full rate equation
  rate = (1.16x10^-6 / sec) * [CH3CO2H]